Abstract theorems

LetX,Y be Banach spaces. Our aim is to develop a general method which will enable us to solve equations of the type

F(u,λ)=v, (2.3)

whereF:X× →Y is a prescribed sufficiently smooth function andv∈Y is given. Theorem 2.1. LetX,Y be real Banach spaces and let (u0,λ0) ∈ X× . Consider aC1_- mappingF=F(u,λ) :X× →Y such that the following conditions hold:

(i) F(u0,λ0)=0;

(ii) the linear mappingFu(u0,λ0) :X→Yis bijective.

Then there exists a neighborhoodU0ofu0and a neighborhoodV0ofλ0such that for every

λ∈V0there is a unique elementu(λ)∈U0so thatF(u(λ),λ)=0. Moreover, the mappingV0λ→u(λ) is of classC1.

Proof . Consider the mappingΦ(u,λ) :X× →Y× defined byΦ(u,λ)=(F(u,λ),λ). It is obvious thatΦ∈C1_{. We apply toΦthe inverse function theorem. For this aim, it} remains to verify that the mappingΦ(u0,λ0) :X× →Y× is bijective. Indeed, we have Φu0+tu,λ0+tλ =Fu0+tu,λ0+tλ ,λ0+tλ =Fu0,λ0 +Fu u0,λ0 ·(tu) +Fλ u0,λ0 ·(tλ) +o(1),λ0+tλ . (2.4) It follows that Fu0,λ0 = ⎛ ⎝Fu u0,λ0 Fλ u0,λ0 0 I ⎞ ⎠ _(2.5)

which is a bijective operator, by our hypotheses. Thus, by the inverse function theorem, there exists a neighborhoodUof the point (u0,λ0) and a neighborhoodVof (0,λ) such that the equation

Φ(u,λ)=f,λ0

(2.6) has a unique solution, for every (f,λ)∈ V.Now it is sufficient to take here f =0 and

our conclusion follows.

With a similar proof one can justify the following global version of the implicit function theorem.

Theorem 2.2. AssumeF:X× →Y is aC1_{-function onX× satisfying} (i) F(0, 0)=0,

(ii) the linear mappingFu(0, 0) :X→Y is bijective.

Then there exist an open neighborhoodI of 0 and aC1_{mappingI λ → u(λ) such that}

u(0)=0 andF(u(λ),λ)=0.

The following result will be of particular importance in the next applications. Theorem 2.3. Assume the same hypotheses onF as inTheorem 2.2. Then there exists an open and maximal intervalI containing the origin and there exists a uniqueC1_-mapping

Iλ→u(λ) such that the following hold: (a) F(u(λ),λ)=0, for everyλ∈I;

(b) the linear mappingFu(u(λ),λ) is bijective, for anyλ∈I; (c) u(0)=0.

Proof . Letu1,u2be solutions and consider the corresponding open intervalsI1andI2on which these solutions exist, respectively. It follows thatu1(0)=u2(0)=0 and

Fu1(λ),λ=0, for everyλ∈I1,

Fu2(λ),λ=0, for everyλ∈I2.

(2.7)

Moreover, the mappings Fu(u1(λ),λ) andFu(u2(λ),λ) are one-to-one and onto onI1, respectively.I2. But, forλsufficiently close to 0 we haveu1(λ)=u2(λ). We wish to show that we have global uniqueness. For this aim, let

I=λ∈I1∩I2; u1(λ)=u2(λ) . (2.8) Our aim is to show thatI =I1∩I2. We first observe that 0∈ I, soI = ∅. A standard argument then shows thatI is closed inI1∩I2. In order to show thatI = I1∩I2, it is sufficient now to prove thatIis an open set inI1∩I2. The proof of this statement follows by applyingTheorem 2.1forλinstead of 0. Thus,I=I1∩I2.

Now, in order to justify the existence of a maximal intervalI, we consider theC1_- curvesun(λ) defined on the corresponding open intervalsIn, such that 0∈In,un(0)=u0,

F(un(λ),λ) = 0 andFu(un(λ),λ) is an isomorphism, for anyλ ∈ In. Now a standard argument enables us to construct a maximal solution on the set∪nIn.

Corollary 2.4. LetX,Ybe Banach spaces and letF:X→Ybe aC1_{-function. Assume that} the linear mappingFu(u) :X→Yis bijective, for everyu∈Xand there existsC >0 such that(Fu(u))−1 ≤C, for anyu∈X. ThenFis onto.

Proof . Assume, without loss of generality, that F(0) = 0 and fix arbitrarily f ∈ Y. Consider the operatorF(u,λ)=F(u)−λ f, defined onX× . Then, byTheorem 2.3, there

exists aC1_{-functionu(λ) which is defined on a maximal intervalIsuch thatF(u(λ)) =}

λ f. In particular,u:=u(1) is a solution of the equationF(u)= f. We assert thatI= . Indeed, we have uλ(λ)= Fu(u) −1 f, (2.9)

souis a Lipschitz map onI, which impliesI= .

The implicit function theorem is used to solve equations of the typeF(u)= f, where

F ∈C1_{(X,Y). A simple method for proving thatFis onto, that is, ImF =Y is to prove} the following:

(i) ImFis open; (ii) ImFis closed.

For showing (i), usually we use the inverse function theorem, more exactly, ifFu(u) is one-to-one, for everyu∈X, then (i) holds. A sufficient condition for that (ii) holds is

thatFis a proper map.

Another variant of the implicit function theorem is stated in the following result. Theorem 2.5. Let F(u,λ) be a C1_{-mapping in a neighborhood of (0, 0) and such that}

F(0, 0)=0. Assume that (i) ImFu(0, 0)=Y;

(ii) the spaceX1:=KerFu(0, 0) has a closed complementX2.

Then there existB1= {u1∈X;u1< δ},B2= {λ∈ ;|λ|< r},B3= {g∈Y;g<

R}, and a neighborhoodUof the origin inX2such that, for anyu1∈B1,λ∈B2, andg∈B3, there exists a unique solutionu2=ϕ(u1,λ,g)∈Uof the equation

Fu1+ϕ

u1,λ,g,λ=g. (2.10)

Proof . LetΓ= X× ×Y, that is, every elementν ∈ Γhas the formν = (u1,λ,g). It remains to apply then implicit function theorem to the mappingG:X×Γ→Y which is

defined byG(u2,ν)=F(u1+u2,λ)−g.

We conclude this section with the following elementary example: letΩbe a smooth bounded domain in N _{and letg be aC}1 _{real function defined on a neighborhood of 0} and such thatg(0)=0. Consider the problem

−Δu=g(u) + f(x) inΩ

u=0 on∂Ω. (2.11)

Assume thatg(0) is not a real number of−ΔinH01(Ω), sayg(0)≤0. If f is sufficiently small, then the problem (2.11) has a unique solution, by the implicit function theorem. Indeed, it is enough to apply Theorem 2.1 to the operator F(u) = −Δu−g(u) and

after observing thatFu(0)= −Δ−g(0). There are at least two distinct possibilities for definingF:

(i) F:C02,α(Ω)→C0α(Ω), for someα∈(0, 1), or (ii) F : W2,p_(Ω)∩W1,p

0 (Ω) → Lp(Ω); In order to obtain classical solutions (by a standard bootstrap argument that we will describe later), it is sufficient to choosep > N/2.